My law schools shady response to mathematical improbability. I'm sure this can be figured out easily but I'm a law guy and never took probabilities in school. Recently my school had a competition for moot court. The results of the semi finals are the only ones in dispute. While other students complained about the outcome discussed below, I'm more curious in the probability that things happened the way they did. Here are the facts: 
1) There were (48) students. 
2) We each had to give give (2) oral arguments. (Argue one side of the case, and then the opposite viewpoint)
3) there were (12) classrooms were arguments were given. 
4) after each person argued in one classroom, the were assigned a different room to give the second and opposing view as mentioned above. 
5) I'm not sure if this is statistically relevant but there were three other students in each room. Essentially all 48 students have their arguments in the 12 different classrooms (randomly assigned), then each of us went to our second randomly assigned classroom. 
6) there was allegedly a curve applied to compensate for the difference in the way the judges graded but since that's what's in dispute, let's assume no curve was used. 
7) the results were that 7 out of the 8 semi-finalists all argued in one particular room. Let's call that the winners room. Keep in mind, these 7 argued in the winners room as well as one of the other 11 rooms. 
My question is, notwithstanding the results, what was the probability that the results would have come out that way if grading was identical. The extent of my math is that each person had a 1/6 chance of moving on to the semifinals, while the chances a winner would come from any given room is less than one. 
 A: There were exactly eight students in classroom 1. 
The probability that the first 7 are among the eight semi finalists is (8/48) * (7/47) * (6/46) * ... (2 / 42): The first is one of eight semi finalists out of 48 students. The second is one of seven remaining semi finalists out of 47 students and so on. That chance is one in 9,203,634. 
The chance that any 7 (not just the first seven) are among the eight semi finalists is 8 times higher, one in 1,150,454 because there are eight in the room that might be the one non-semi finalist. 
The chance of 7 semi finalists in one of 12 rooms is at most 12 times higher (a tiny bit less because of the tiny chance that there are two rooms with 7 semi finalists), that is less than one in 95,871. Adding the chance for "8 finalists in the same room" makes very little difference. 
The chances that this happened in one specific "moot court" by coincidence are suspiciously low. On the other hand, I don't know how many law schools there are and how many such "moot court" decisions they hold, so the chance that this would happen somewhere at some time might be reasonably high. 
A: Let's suppose you randomly assign $8$ students  to be semifinalists (the other $40$ losers are not relevant; whether you do this before or after the speeches is also irrelevant), and you randomly (and independently) assign each student two different rooms of the $12$ rooms.  
EDIT: This is not exactly how it was done, rather the assignment was made so that each room had the same number of students for both first and second speeches, but it should make little difference.
The probability that a given student is assigned a particular room (say room #1) is $2/12 = 1/6$.  
EDIT: This part is true regardless of how the assignment was done, as long as no room is favoured over another.  A given student's probability of being assigned a given room is the same as for each other room, and the total must be $2$ because exactly $2$ rooms will be assigned to the student.
Thus the probability that $7$ or $8$ of the semifinalists are assigned to  room #1
is $(1/6)^8 + 8 (5/6)(1/6)^7 = 41/1679616$.  The probability that there is at least one room that $7$ or $8$ of the semifinalists are assigned to is (very slightly) less than $12 \times 41/1679616 = 41/139968 \approx 0.00029$.  
